The drawing shows two boxes resting on frictionless ramps. One box is relatively

Question

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 8 and 43 kg. If A and B are 3.5 and 1.0 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?

Explanation / Answer

The height, h, of A above B is 2.5 m. Let M1 and M2 be the masses of the lighter and heavier boxes. At point A, the gravitational potential energies with respect to point B of the 2 boxes are GPE1 = M1*g*h GPE2 = M2*g*h At point B, the kinetic energies of the 2 boxes are KE1 = (1/2)*M1*V1^2 KE2 = (1/2)*M2*V2^2 a&b) At point B, those GPEs will be 100% converted to Kinetic Energies. So M1*g*h = (1/2)*M1*V1^2 V1=sqrt(2gh)=7m/sec M2*g*h = (1/2)*M2*V2^2 V12=sqrt(2gh)=7m/sec c) KE2/KE1=M2/M1=43/8=5.375

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